Cascode side move looks right, but other side of 820p now connects
Q13 Q10 Q9 instead of Q4 Q6. Subtle reason for this, or simple oops?

Oops!

Quote:

Elvee, could you explain a little how the bias servo operates? I don't completely follow how the network of Q2, Q7, Q12, and Q13 works

In fact those transistors form an "AND" gate but in the analog domain: for Q2 to conduct (and thus to reduce the quiescent current via the common mode loop), a voltage has to be present across R8 (i.e. a current must flow into Q8) AND the compensating transistor Q7 has to be biased, which requires Q13 to conduct, and therefore ultimately Q10 too (Q12 is just a compensating transistor).

The subtlety of the circuit is that it doesn't simply work digitally, but in an analog manner too: if the current in one of the output transistor is lower, this can be compensated within certain limits by an increase in the other one.

The result is that the currents can never fall below a minimum of 80mA, and vary in a complementary manner.

This also forces both OP transistors to always remain active, and I mean really active: it isn't simply a "passive" bias current superimposed on the working current: not only is each transistor active, but so is the whole upstream circuitry, up to and including the input stage.

This means that the OP transistors play no role in deciding the amount or the repartition of the quiescent current, nor the output current's value.
This is decided at a signal level, and is the result of the "negociation" taking place between the two control loops: the "noble" differential one and the ancillary common mode one.
Here is the schematic again for an easier reference.

The result is that the currents can never fall below a minimum of 80mA, and vary in a complementary manner.

This also forces both OP transistors to always remain active, and I mean really active: it isn't simply a "passive" bias current superimposed on the working current: not only is each transistor active, but so is the whole upstream circuitry, up to and including the input stage.

This means that the OP transistors play no role in deciding the amount or the repartition of the quiescent current, nor the output current's value.
This is decided at a signal level, and is the result of the "negociation" taking place between the two control loops: the "noble" differential one and the ancillary common mode one.

And why are we the only ones currently proposing circuits with
two such simultaneously active cooperative feedback loops?

And why are we the only ones currently proposing circuits with
two such simultaneously active cooperative feedback loops?

I think there are two main reasons: one has to do with tradition/habits/fear of the unknown, etc, and the other is a (partially) justified distrust of something (felt like) over-sophisticated, over-complicated and too delicate to work in practice.

Was messin with your front end and finals grafted with a centralized quiescent control.
I come to realize both loops need independent compensation for stability. But I have
not yet mastered how to make (two schottkys, resistor, transistor) completely stable.
I return a control current to the 330R at the emitter of whats Q1 in your drawing...
Base of Q1 is simply biased by two diodes and 1mA. Works fine up to about 8K, but
anything above makes the whole sim go completely nuts...

Wasn't ever stable till I axed one stage of drivers.
But then I had to flip the whole front upsidedown.
Had no idea what I was doing, random experiment.

Stability still goes bannanas whenever it clips.
Recovery from clipping here is just plain awfull.
I would soft clip this with Zeners across R11.

One 2n3055 gets an assist from base current, the
other dumps base current into the negative rail.
So the collector currents will never be symmetrical.
Took me a while to realize I should stop obsessing
for a perfect current balance, not posssible...